I am reading that there exists a unitary matrix $C$ (the charge conjugation) matrix such that each matrix $C\Gamma^{A}$ is either symmetric or anti-symmetric. Now, $\Gamma^{A} = \{ {\bf 1}, \gamma^{\mu}, \gamma^{\mu \nu}, \ldots \}$ are the basis of Clifford algebra in some even dimension. Then I read that symmetry only depends on the rank $r$ of the matrix $\Gamma^A$ so they write: $$ (C\Gamma^{(r)})^T = -t_r C\Gamma^{(r)}, \,\,\,\,\,\,\, t_r=\pm1 \tag{1}$$ where this $\Gamma^{(r)}$ matrix belongs to the set I gave above.

How do they get (1)? I really do not get it.

Then they say that for rank $r=0$ and $r=1$ they obtain $$C^T = -t_0C, \,\,\,\,\,\,\, (\gamma^{\mu})^T=t_0t_1C\gamma^{\mu}C^{-1} \tag{2}$$

Again how do they obtain (2), especially the right hand side equation? Why they consider the transpose of the matrix? Finally they comment that the previous relations suffice to determine the symmetries of all $C\gamma^{\mu \nu \ldots}$ and thus all coefficients. E.g. $t_2=-t_0$ and $t_3 = -t_1$, and generally $t_{r+4}=t_r$. Again, why?

  • $\begingroup$ $\uparrow$ Reading where? $\endgroup$ – Qmechanic Jan 4 '15 at 8:38

I'll have to lots of assumptions about your terminology, so I'll try to state each time I use an assumptions about your terminology so you know exactly when to protest if I guess wrong.

So first, we can assume this C, or we can derive it. We will start by assuming it, and assuming that it satisfies $(CM)^T=-t_rCM$ for every $M$ in your additive basis $B=\{1,\gamma^\mu, ..., \gamma^{\mu\nu}, ...\}$ where $t_r$ depends only on the rank (I'll assume by rank they mean how many indicies are used to label the basis vector, not the rank as a matrix in some representation).

So $r=0$ has only one additive basis element, $1$, and so our equation gives $(C1)^T=-t_0C1$, so $C^T=-t_0C$ as asked.

Now consider $r=1$, there are $n$ additive basis elements where $n$ is even because you said so, but really where $n$ is the size of $\{\gamma^\mu, ... \gamma^\nu, ...\}$, so let $\gamma^\mu$ be an arbitrary one of them, then

$(C\gamma^\mu)^T=-t_1C\gamma^\mu$ is our assumption. Assuming we have a matrix representation of the element $\gamma^\mu$ of the Clifford Algebra, then we have a normal matrix product and a transpose, so we get $(\gamma^\mu)^TC^T=-t_1C\gamma^\mu$ from the general matrix rule $(AB)^T=B^TA^T$.

Next we use the previous result that $C^T=-t_0C$ to get: $(\gamma^\mu)^T(-t_0C)=-t_1C\gamma^\mu.$

For the next step we multiply both sides (on the right) by $-t_0C^{-1}$ and get: $(\gamma^\mu)^T(-t_0C)(-t_0C^{-1})=-t_1C\gamma^\mu(-t_0C^{-1}).$ And move some scalars around to get $(\gamma^\mu )^T(t_0^2CC^{-1})=t_0t_1C\gamma^\mu C^{-1}.$

And since $t_0$ is either +1 or -1, its square is 1, and $CC^{-1}$ is the identity, so we have: $(\gamma^\mu )^T=t_0t_1C\gamma^\mu C^{-1}$. So the assumptions gives us the identities you listed, we haven't even used the assumption that $C$ is unitary for these identities so far.

Now for the other identities, you say you want to show the "other symmetries", and I'm not sure what that is, but lets see what we can show. Look at a higher rank additive basis element such as $\gamma^{\mu\nu}$, we know that

$(C\gamma^{\mu\nu})^T=-t_2C\gamma^{\mu\nu}$, so $(\gamma^\nu)^T(\gamma^\mu)^TC^T=-t_2C\gamma^{\mu\nu}.$

And we know the transpose of those gamma matrices from our previous result for $t_1$, so we'll put those in and get: $(t_0t_1C\gamma^\nu C^{-1})(t_0t_1C\gamma^\mu C^{-1})C^T=-t_2C\gamma^{\mu\nu}.$

We can move the scalars on the right hand side and combine the $C^{-1}C$ to get:

$t_0^2t_1^2C\gamma^\nu \gamma^\mu C^{-1}C^T=-t_2C\gamma^{\mu\nu}.$ Again the $t_r$ square to 1 and we can replace $C^T$ with the $t_0$ identity (the first one we showed, i.e. $C^T=-t_0C$) to get:

$C\gamma^\nu \gamma^\mu C^{-1}(-t_0C)=-t_2C\gamma^{\mu\nu}.$

And now we finally use the fact that this is a Clifford Algebra for the first time, and note that $\gamma^{\mu\nu}$ is the name for the result of the multiplicative product $\gamma^\mu \gamma^\nu$, and that we assume the gamma matrices to be orthogonal i.e. $\gamma^\mu \gamma^\nu +\gamma^\nu \gamma^\mu=0$.

So we get to turn $C\gamma^\nu \gamma^\mu C^{-1}(-t_0C)=-t_2C\gamma^{\mu\nu}$ into the following sequence:

$C\gamma^\nu \gamma^\mu C^{-1}(-t_0C)=-t_2C\gamma^{\mu}\gamma^{\nu},$

$C\gamma^\nu \gamma^\mu C^{-1}(-t_0C)=t_2C\gamma^{\nu}\gamma^{\mu},$

$C\gamma^\nu \gamma^\mu (-t_0)=t_2C\gamma^{\nu}\gamma^{\mu},$

$\gamma^\nu \gamma^\mu (-t_0)=t_2\gamma^{\nu}\gamma^{\mu},$


Where in the second to last step we used that $C$ was unitary for the first time (actually we just needed that it was invertible). In the last step we used that gamma matrices are invertible, up to a scale they are their own inverse, and the scale and be dropped too so you can multiple both sides on the left by $\gamma^\mu \gamma^\nu C^T$ to do it in one step (well a second step to remove the identical scalars on both sides.

Rather than do al the steps for $t_3$, just note that when you replace every transpose of a gama with the $t_1$ identity you'll et three factors of $t_0$ and three factors of $t_1$ on the left hand side originally, but the $C^T$ will use up one of your $t_0$ so really you'll just have a relationship between $t_1$ and $t_3$ and this time you'll have three orthogonal gamma matrices to commute past each other so you will still the sign switching from $\gamma^\alpha\gamma^\beta\gamma^\delta$ to $\gamma^\delta\gamma^\beta\gamma^\alpha$ (but when you have four you won't because the first commutes past 3, then the second past 2 then the last past each other so 6 sign flips total).

We haven't showed existence, but the identity matrix works for $C$ with $t_0=t_1=1$, so its possible even without our assumption (still unused) that the dimension of the gamma matrices is even (obviously the dimension of the algebra as a linear space is even, it is a power of two). Maybe they mean that you can select $t_0$ and $t_1$ at will (which then determine the rest) and get a $C$. Clearly the transposes of the gammas are still orthogonal in that their symmetric product yields zero, and multiplying by $t_0t_1$ doesn't change the size so if you want to find a unitary similarity between them, that might be all they want.

  • $\begingroup$ Why you say that $\gamma^{\mu} \gamma^{\nu} + \gamma^{\nu} \gamma^{\nu} = 0$? The Clifford algebra says that $\gamma^{\mu} \gamma^{\nu} + \gamma^{\nu} \gamma^{\nu} = 2\eta^{\mu \nu}$ $\endgroup$ – Marion Jan 4 '15 at 14:58
  • $\begingroup$ I assume the n vectors in the basis make an orthonormal set, so $\gamma^\mu\gamma^\mu=\pm 1$, and for two different indices we get $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=0$ (e.g. $\eta$ is diagonal). It's only when you take all different indices that you get a new element for your basis. We already have 1 in the additive basis so $\gamma^\mu\gamma^\mu=\pm 1$ wouldn't be linearly independent. Every rank $r$ basis element needs $r$ different indices, if there were repetition, then you could anticommute until they are next to each other then combine to get a scalar and then be lower rank. $\endgroup$ – Timaeus Jan 4 '15 at 17:24
  • $\begingroup$ Thanks, I am struggling to understand why we choose that $\mu \neq \nu$. I know that in this case what you have written is right, I am not sure yet why you say that $\gamma^{\mu \mu}$ is not an element in my basis. $\endgroup$ – Marion Jan 6 '15 at 17:01
  • $\begingroup$ You have an additive basis, in the sense that anything in the algebra can be written (uniquely) as a linear combination of your additive basis elements. So you'd have $1,\gamma^1,\gamma^2$, and $\gamma^1\gamma^2$, but since $\gamma^1\gamma^1=\pm 1$ if you put it in your set the set would no longer be linearly independent. Effectively $\gamma^1\gamma^1$ is rank 0 (because it is a scalar) rather than rank 2. $\endgroup$ – Timaeus Jan 7 '15 at 17:56

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